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I was going through the Theorem 2 :

Suppose $u(x)$ represents agents preferences, $\succsim$ and $f : \mathbb{R} \rightarrow \mathbb{R} $ is a strictly increasing function. Then the new new utility function $v(x) = f(u(x))$ also represents the agent's preferences $\succsim$. So mathematically: $v(x) \geq v(y) \iff x \succsim y$

I am having trouble understanding this above double implication statement.

Suppose $f$ is not strictly increasing (i.e just a non decreasing function) and if we do monotonic (not strictly) transformation of the utility (like this) $v : z \mapsto 0$ then for two alternatives $x,y$ let $x \succ y$ but their utility values $v(u(x)) = v(u(y))$ are same. So shouldn't we consider it satisfying the equivalence relation above Since for $x \succsim y$ we have $x\succ y$ true and for $u(x) \geq u(y)$ we have $u(x) = u(y)$ true making the equivalence hold true for non decreasing function also.

I know this is wrong , since the ordering is not maintained after the transformation but Im not able to understand it in the form of mathematical statement of double implication, what am I doing wrong here.

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Since $u(x)$ represents $\succsim$, $\succsim$ must be rational. By completeness, $\neg (x\succsim y) \iff x \prec y$. Therefore, the claim that $$v(x) \geq v(y) \iff x\succsim y, \forall x, y$$ is equivalent to its contrapositive (for both sides): $$v(x) < v(y) \iff x \prec y, \forall x, y$$ Pick any $x\prec y$, we should have $u(x) < u(y)$. However, suppose your transformation $f$ is constant (non-decreasing, but not strictly increasing), $f(u(x)) = f(u(y))$. That is, not $v(x) < v(y)$, which means $v(\cdot)$ does not represent $\succsim$.

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  • $\begingroup$ Ok just to be clear if we write down the preference relation set $\succsim$ where $x \succsim y $, iff $(x,y) \in$ $\succsim$, so for $x \succ y$ to be true it should be that $(x,y) \in$ $\succsim$ and $(y,x) \not\in$ $\succsim$ $\endgroup$
    – hr08
    Commented Aug 24 at 5:21
  • $\begingroup$ also what i have learned is $x \succ y \iff (x \succsim y) \wedge \neg(y \succsim x)$, so I didnt understand the negation of weak preference $\endgroup$
    – hr08
    Commented Aug 24 at 10:15
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    $\begingroup$ If I understand correctly, your two comments are saying the same thing. You are correct, what you mentioned was the definition of a weak preference without any premise. However, I was using a property here: if $\succsim$ can be represented by $u(\cdot)$, $\succsim$ must be complete, which means for any $x, y$: $\neg(x\succsim y) \iff x \prec y$. $\endgroup$
    – DiZ
    Commented Aug 24 at 20:32
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    $\begingroup$ It seems like Simon's lecture notes did not mention this property, but I believe it is easy to verify yourself. The idea is that $>$ possesses transitivity and completeness, so does $\succsim$. $\endgroup$
    – DiZ
    Commented Aug 24 at 20:37
  • $\begingroup$ Thanks alot got it. $\endgroup$
    – hr08
    Commented Aug 25 at 3:44

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